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JEE MainPhysicsDual Nature of Matter

A metallic surface is illuminated with light of wavelength , and the minimum de Broglie wavelength of the emitted photoelectrons is found to be _d . When the wavelength of the incident light is changed to 2 , the minimum de Broglie wavelength of the emitted photoelectrons becomes _d 2 . The threshold wavelength for this metallic surface is:

Options

  1. A6 7
  2. B3 7
  3. C3
  4. D3 2

Correct answer

D. 3 2

Step-by-step solution

The de Broglie wavelength of an electron is related to its kinetic energy K by the relation: _d = h 2mK K = h^2 2m _d^2 When the minimum de Broglie wavelength becomes _d 2 , the new maximum kinetic energy K' becomes: K' = h^2 2m ( _d 2 )^2 = 4 ( h^2 2m _d^2 ) = 4K Using Einstein's photoelectric equation for the first case (incident wavelength ): K = hc - hc ₀ For the second case (incident wavelength 2 ): 4K = hc ( 2 ) - hc ₀ = 2hc - hc ₀ Substitute K from the first equation into the second: 4 ( hc - hc ₀ ) = 2hc -

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